Find the no. of trailing zeroes of $20190!!$ Find the no. of trailing zeroes of $20190!!$
What I did was I found the no. of trailing zeroes of $20190!$ and divided it by $2$ and got $2522$, but the correct answer was $2521$. What's the correct solution here?
 A: The answer by liaombro already mentioned this,
but did not go into details,
so I would like to explain it a little bit here. 

Here, $n!!$ is taken to denote the double factorial of $n$.
By definition,
\begin{align*}
  20190!! &= (20190) (20188) (20186) \dotsm (2) \\
          &= 2(10095) \cdot 2(10094) \cdot 2(10093) \dotsm 2(1) \\
          &= 2^{10095} \cdot (10095)(10094)(10093) \dotsm (1) \\
          &= 2^{10095} \cdot 10095!.
\end{align*}
Now you should be able to figure out the number of trailing zeros of $10095$
by yourself.
After that, note that $10095!$ contains strictly less factors of $5$ than $2$.
Therefore, multiplying $10095!$ by $2^{10095}$ does not introduce new trailing zeros.
A: As per the other answers, $20190!!=2^{10095}\cdot 10095!$
Now, according to Legendre's Theorem
$$\nu_p(n!)=\frac{n-s_p(n)}{p-1}$$
With $p=2$
$$\nu_2(10095!)=\frac{10095-s_2(10011101101111)}{2-1}=10095-10=10085$$
With $p=5$
$$\nu_5(10095!)=\frac{10095-s_5(310340)}{5-1}=\frac{10095-11}{4}=2521$$
and finally
$$20190!!=2^{10095+10085}\cdot 5^{\color{red}{2521}} \cdot Q$$
where $Q$ is not contributing to the trailing zero's, since it doesn't contain any $2$ or $5$ in its factorization.
A: Hint: $20190!! = 2^{10095} \cdot  10095!$
